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HSC 2013 MX2 Marathon (archive) (1 Viewer)

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Sy123

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Re: HSC 2013 4U Marathon

Hope you don't mind me expanding on this:








I have written up a solution, but I will do what you are doing and wait until night and post my solution after yours.
 

jeffwu95

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Re: HSC 2013 4U Marathon

its amazing how much you can forget in a month
 

Sy123

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Re: HSC 2013 4U Marathon

Here is another question:



















 
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seanieg89

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Re: HSC 2013 4U Marathon

Just a point on nomenclature: the object in d) is called a formal power series, not a polynomial.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Also Sy, you made a typo when you defined the sequence. No biggie, just people unfamiliar with the Fibonacci sequence might define it using the typo.

The part, should be "n-2".
 

Sy123

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Re: HSC 2013 4U Marathon

Also Sy, you made a typo when you defined the sequence. No biggie, just people unfamiliar with the Fibonacci sequence might define it using the typo.

The part, should be "n-2".
Oops, well I'll fix that now. Thanks
 

Sy123

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Re: HSC 2013 4U Marathon

Hope you don't mind me expanding on this:







iii) Not sure how to actually deduce it from previous results but I can easily evaluate both sides.

LHS is a telescoping sum from which we get:



RHS is a limiting sum of a geometric series:



iv)

First term of LHS -> 1/2
First term of RHS -> 1/2

v) k=5 -> 5(5+1)=30 =/=2^5=32
Therefore for k=5 the non-equality holds

vi) Note how k=1 is a solution, when we see k=2
For k(k+1)=2^k

2(3) > 2^2
6 > 4

However we observed before that k=5

30 < 32

Hence there must exist a solution in between k=2 and k=5 whereby the transition from LHS > RHS and LHS < RHS takes place.


Not sure if this is the intended solution. Also I will post solution to Fibonacci question tomorrow night if no-one posts a solution by then.
 

Sy123

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Re: HSC 2013 4U Marathon

Something a little easier:





 

RealiseNothing

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Re: HSC 2013 4U Marathon

iii) Not sure how to actually deduce it from previous results but I can easily evaluate both sides.

LHS is a telescoping sum from which we get:



RHS is a limiting sum of a geometric series:



iv)

First term of LHS -> 1/2
First term of RHS -> 1/2

v) k=5 -> 5(5+1)=30 =/=2^5=32
Therefore for k=5 the non-equality holds

vi) Note how k=1 is a solution, when we see k=2
For k(k+1)=2^k

2(3) > 2^2
6 > 4

However we observed before that k=5

30 < 32

Hence there must exist a solution in between k=2 and k=5 whereby the transition from LHS > RHS and LHS < RHS takes place.


Not sure if this is the intended solution. Also I will post solution to Fibonacci question tomorrow night if no-one posts a solution by then.
I should have put something like "without substitution of further k values, deduce that there is a positive real solution besides k=1".

The intended solution was to see that for some value k (ie you did k=5), the sums are not equal and hence the ratios are not equal (one is larger).

But as you approach infinity the sums become equal, and hence the magnitudes of the ratios of the series must switch so that the other is larger.

If the ratios cross over, then there must be a positive real solution.

Not sure if this makes sense since it's 3am, but w/e.
 

Sy123

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Re: HSC 2013 4U Marathon

I should have put something like "without substitution of further k values, deduce that there is a positive real solution besides k=1".

The intended solution was to see that for some value k (ie you did k=5), the sums are not equal and hence the ratios are not equal (one is larger).

But as you approach infinity the sums become equal, and hence the magnitudes of the ratios of the series must switch so that the other is larger.

If the ratios cross over, then there must be a positive real solution.

Not sure if this makes sense since it's 3am, but w/e.
Oh I see, nice solution.
 

HeroicPandas

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Re: HSC 2013 4U Marathon

Just in case other people did not see it.
Test:
S(-1) = 0

Therefore x=-1 is a root

Since roots are in an arithmetic sequence, let roots be



(d+1)(d-1)=-1
d^2=0
So d=0....

Therefore triple root at x = -1? (or roots are x = -1, -1, -1)
 

Sy123

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Re: HSC 2013 4U Marathon

Test:
S(-1) = 0

Therefore x=-1 is a root

Since roots are in an arithmetic sequence, let roots be



(d+1)(d-1)=-1
d^2=0
So d=0....

Therefore triple root at x = -1? (or roots are x = -1, -1, -1)
That is one case of it, yes.

I should of extended the question to:



So yeah consider if -1 is the first or the last term etc.
 
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